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There are  members in the Trumpington marching band, and when they line up in rows of 26, there are 4 band members left over. If  is an integer and there are fewer than 1000 band members, what is the maximum number of people that could be in the Trumpington marching band?       

I don't know how to work modulus problems,

I don't think they taught it when I was in school.   

In fact, I never even heard of it until I saw it in this website. 

So I'm going to kind of wing it.     

There are fewer than 1000 members,  

so start with 999.   

Divide 999 by 26 and see what the answer is.   

                                                                                         999 / 26  =  38.423   

Take the whole number part of that quotient, 

which is 38, and multiply it by 26. 

                                                                                           38 • 26  =  988   

So 988 is the number of band members that will  

fill up 38 rows of 26 members per row with none    

left over.  We need 4 left over, so just add it on.    

                                                                                          988 + 4  =  992   

And there you have it.  992 band members is the most,   

short of 1000, that will fit into rows of 26 members per row,     

and have 4 musicians left over.  Probably the piano players;   

those Steinways are hard to trundle around on a football field.

_.

n=7

Expand   (4x^2 - 7y)^7

16384 x^14 - 200704 x^12 y + 1053696 x^10 y^2 - 3073280 x^8 y^3 + 5378240 x^6 y^4 - 5647152 x^4 y^5 + 3294172 x^2 y^6 - 823543 y^7
 

2889 is the correct answer !!

364 is the correct answer !!

First, you have to count all the digits between 1 and 999.

1 to 9 ==9 digits

10 to 99 ==90 x 2 ==180 digits

100 to 999 ==900 x 3 ==2,700 digits

Total =9 + 180 + 2700 ==2,889 digits between 1 and 999

There are: 300 "sevens" between 1 and 999.

Therefore, the probability of picking a "7" is:

300 / 2,889 ==100 / 963

Simplify 2 * (9)^(1/3) + 9^(1/3) + 9^(1/3)

==2 * 3^(2/3) + (1 * 3^(2/3)) + (1 * 3^(2/3))

==[2 + 1 + 1] * 3^(2/3)

==4 * 3^(2/3)  [this is the same as:]

==4 * 9^(1/3)

Cube roots mean that it takes three numbers to get that one number

We can solve this problem by casework.

Case 1: No siblings sit together in the same row.

In this case, we can seat the siblings in any order. There are 6! ways to seat 6 people in 6 chairs, so there are 6! ways to seat the siblings in this case.

Case 2: One sibling pair sits together in the same row.

There are 2 ways to choose which sibling pair sits together. Once we have chosen the pair, we can seat them in 2 ways (AB or BA). Then, we can seat the remaining 4 people in 4! ways. So, there are 2⋅2⋅4!=48 ways to seat the siblings in this case.

Case 3: Two sibling pairs sit together in the same row.

There are (23​)=3 ways to choose which two sibling pairs sit together. Once we have chosen the pairs, we can seat them in 3 ways (AABB, ABAB, or BABA). Then, we can seat the remaining 2 people in 2! ways. So, there are 3⋅3⋅2!=18 ways to seat the siblings in this case.

Total:

The total number of ways to seat the siblings is 6! + 48 + 18 = 786.

Because these are cube roots so it's cube root of 9 

We can think of Anna and Bob as a single unit, so there are 8 units that need to be arranged around the table. This can be done in 7! ways. However, we need to divide by 2 since rotations of the arrangement are the same. So, the number of ways to arrange 9 friends around a circular table if two of them insist on having exactly two people seated between them is 7!/2 = 2520.

After each step, Franklin's position changes by an even number. After 12 steps, the sum of these changes must also be even. Therefore, Franklin could end up at any point whose coordinates are both even or both odd. There are 11 even integers and 11 odd integers from 0 to 12, so there are 112=121​ possible points that Franklin could end up at.

Here is a proof by cases:

Case 1: Franklin ends up at a point with both even coordinates. There are 11 even integers from 0 to 12, so there are 11⋅11=121 possible points in this case.

Case 2: Franklin ends up at a point with both odd coordinates. There are 11 odd integers from 0 to 12, so there are 11⋅11=121 possible points in this case.

Therefore, there are a total of 121+121=242 possible points that Franklin could end up at.

Thanks, that's right!

Thanks, that's right!

look here

https://web2.0calc.com/questions/help-i-need-help-with-percent

Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 21 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?   

Call the smallest integer x.  Then each following integer  

will be 1 larger than the integer that preceded it.  

x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)  =  21 • (x + 6)  

                                                                          7x + 21  =  21x + 126  

Combine like terms                                                –14x  =  105   

Divide both sides by –14                                             x  =  –7.5  

The problem fails already, since 7.5 is not  

an integer, whether it's positive or negative.            NO SOLUTION  

But let's press ahead and see if this would have worked.   

Check answer by substituting the value into the problem.   

(–7.5) + (–7.5 + 1) + (–7.5 + 2) + (–7.5 + 3) + (–7.5 + 4) + (–7.5 + 5) + (–7.5 + 6)  =  (21)(–7.5 + 6) 

(–7.5) + (–6.5) + (–5.5) + (–4.5) + (–3.5) + (–2.5) + (–1.5)  =  (21)(1.5) 

                                                                                  –31.5  =  –31.5  

So, –7.5 would have been a good answer IF the problem

had not specified that the numbers have to be integers.  

_.

I drove to the beach at a rate of 40 miles per hour.  If I had driven at a rate of 50 miles per hour instead, then I would have arrived 45 minutes later.  How many miles did I drive?  

You mean 45 minutes earlier.  Obviously, if you drive faster, you get there sooner.  

This problem makes use of

the following relationship:                   Distance = Velocity x Time 

                                                           D  =  V • T  

case 1                                                 D  =  (40) • (T)  

case 2                                                 D  =  (50) • (T – 45)  

Since the Distance, D, is the  

same for both cases, let's set       

the "V•T"s equal to each other.             (50)(T – 45)  =  (40)(T)  

                                                               50T – 2250  =  40T  

Subtract 40T from both sides                  10T – 2250  =  0  

Add 2250 to both sides                                       10T  =  2250  

Divide both sides by 10                                           T  =  225   (this is in minutes)  

Divide minutes by 60 to get hours          225 minutes  =  3.75 hours  

Plug this T back into original equation                     D  =  (40 mi/hr) • (3.75 hr)  =  150 miles  

_.

The formula for the sum of the internal angles of a regular polygon is

sum = (n – 2)(180) where n is the number of sides.   

For a pentagram, n is 5         so sum = (3)(180) = 540   

That's the sum, so an individual internal angle is a fifth of that, i.e., 108.  

That makes FED equal 180 – 108 = 72   

The sum of the angles of a triangle is 180, so FDE = 180 – (90 + 72) = 18  

Angle FDE is 18^o   

had to edit because I misspelled polygon, grrr.  

_.

\(\displaystyle f(n) = \frac{4n+\sqrt{4n^{2}-1}}{\sqrt{2n+1}+\sqrt{2n-1}} \\ \displaystyle= \frac{4n+\sqrt{2n+1}.\sqrt{2n-1}} {\sqrt{2n+1}+\sqrt{2n-1}} .\frac{\sqrt{2n+1}-\sqrt{2n-1}}{\sqrt{2n+1}-\sqrt{2n-1}} \\ \displaystyle=\frac{4n\sqrt{2n+1}+(2n+1)\sqrt{2n-1}-4n\sqrt{2n-1}-(2n-1)\sqrt{2n+1}}{(2n+1)-(2n-1)}\\ \displaystyle=\frac{(2n+1)\sqrt{2n+1}-(2n-1)\sqrt{2n-1}}{2}.\)

\(n=1 : f(1)=(3\sqrt{3}-1\sqrt{1})/2, \\ n=2 : f(2)=(5\sqrt{5}-3\sqrt{3})/2, \\ n = 3 : f(3) = (7\sqrt{7}-5\sqrt{5})/2, \\ ............................ \\ ............................\\ n=39 : f(39) = (79\sqrt{79}-77\sqrt{77})/2\\ n = 40 :f(40) =(81\sqrt{81}-79\sqrt{79})/2. \)

\(\displaystyle f(1)+f(2)+\dots+f(40)=-\frac{1\sqrt{1}}{2}+\frac{81\sqrt{81}}{2}=364.\)

Why not: 2 * (3 * 3) + (3*3) +(3*3)=36

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Simplifying the denominators:

1/(a-1) + 1/(b-1) = 1/(1/2 + √5/2 - 2/2) + 1/(1/2 - √5/2 - 2/2)

1/(a-1) + 1/(b-1) = 1/(√5/2 - 3/2) + 1/(-√5/2 - 3/2) <------THIS IS WRONG.

Your answer is bullshit and so is your spam!

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